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A356446
Number of permutations f of {1,...,n} with f(1) = 2 and f(2) = 1 such that the numbers f(k)*f(k+1) (0 < k < n) are distinct and Sum_{k=1..n-1} 1/(f(k)*f(k+1)) = 1.
1
0, 0, 0, 0, 1, 1, 1, 1, 2, 1, 11, 7, 61, 388, 2933, 2741
OFFSET
2,9
COMMENTS
Conjecture: a(n) > 0 for all n > 5.
This is stronger than Conjecture 1 in A322069.
LINKS
Guoniu Han, On the existence of permutations conditioned by certain rational functions, Electron. Res. Arch., 28 (2020), 149-156.
Zhi-Wei Sun, On permutations of {1,...,n} and related topics, J. Algebraic Combin., 2021.
EXAMPLE
a(6) = 1 with the permutation (f(1),...,f(6)) = (2,1,3,4,5,6) meeting the requirement. Note that 2*1 = 2, 1*3 = 3, 3*4 = 12, 4*5 = 6 and 5*6 = 30 are pairwise distinct and 1/2 + 1/3 + 1/12 + 1/20 + 1/30 = 1.
a(9) = 1 with the permutation (f(1),...,f(9)) = (2,1,3,8,7,6,5,9,4) meeting the requirement. Note that 2*1 = 2, 1*3 = 3, 3*8 = 24, 8*7 = 56, 7*6 = 42, 6*5 = 30, 5*9 = 45 and 9*4 = 36 are pairwise distinct and 1/2 + 1/3 + 1/24 + 1/56 + 1/42 + 1/30 + 1/45 + 1/36 = 1.
a(11) = 1 with the permutation (f(1),...,f(11)) = (2,1,4,3,5,11,6,7,8,9,10) meeting the requirement. Note that 2*1 = 2, 1*4 = 4, 4*3 = 12, 3*5 = 15, 5*11 = 55, 11*6 = 66, 6*7 = 42, 7*8 = 56, 8*9 = 72 and 9*10 = 90 are pairwise distinct and 1/2 + 1/4 + 1/12 + 1/15 + 1/55 + 1/66 + 1/42 + 1/56 + 1/72 + 1/90 = 1.
a(12) > 0 since for the permutation (f(1),...,f(12)) = (2,1,4,5,3,6,8,10,11,12,7,9) the 11 numbers 2*1 = 2, 1*4 = 4, 4*5 = 20, 5*3 = 15, 3*6 = 18, 6*8 = 48, 8*10 = 80, 10*11 = 110, 11*12 = 132, 12*7 = 84 and 7*9 = 63 are pairwise distinct and 1/2 + 1/4 + 1/20 + 1/15 + 1/18 + 1/48 + 1/80 + 1/110 + 1/132 + 1/84 + 1/63 = 1. Note that f(3) = 4, f(4) = 3 and f(5) is among 5, 6, 8 for all the other permutations f of {1,...,12} meeting the requirement.
MATHEMATICA
(* A program to find all the permutations f of {1, ..., 10} with f(1) = 2 and f(2) = 1 such that those f(k)*f(k+1) (k = 1..9) are pairwise distinct and Sum_{k=1..9} 1/(f(k)*f(k+1)) = 1. *)
V[i_]:=V[i]=Part[Permutations[{3, 4, 5, 6, 7, 8, 9, 10}], i]
m=0; Do[U={2}; Do[U=Append[U, V[i][[j]]*If[j==1, 1, V[i][[j-1]]]], {j, 1, 8}];
If[Length[Union[U]]==9&&Sum[1/U[[j]], {j, 1, 9}]==1, m=m+1; Print[m, " ", V[i], " ", U]]; Label[aa], {i, 1, 8!}]
CROSSREFS
KEYWORD
nonn,more
AUTHOR
Zhi-Wei Sun, Aug 07 2022
EXTENSIONS
a(13)-a(15) from Jinyuan Wang, Aug 09 2022
a(16)-a(17) from Hugo Pfoertner, Aug 19 2022
STATUS
approved